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Q.

The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If the temperature of source is kept constant and its efficiency is to be raised to 60% then the required temperature of sink will be

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a

600 K

b

500 K

c

400 K

d

100 K

answer is C.

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Detailed Solution

1−T2T1=50100⇒T1=2T2=2×500k=1000k∴1−T211000=60100⇒T21=0.4×1000k=400K
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